Problem: Evaluate the definite integral. $\int^{3}_{1}\left(\dfrac{15+2x^3}{x^4}\right)\,dx = $ Choose 1 answer: Choose 1 answer: (Choice A) A $\ln\left(\dfrac13\right)-\dfrac{130}{27}$ (Choice B) B $\dfrac{40}{5}$ (Choice C) C $2\ln\left(3\right)+\dfrac{130}{27}$ (Choice D) D None of the above
Solution: First, simplify and use the power and natural log rules: $\begin{aligned}\int^{3}_{1}\left(\dfrac{15+2x^3}{x^4}\right)\,dx ~&=~\int^{3}_{1}\left(\dfrac{15}{x^4}+\dfrac{2x^3}{x^4}\right)\,dx \\&=~\int^{3}_{1}\left(15x^{-4}+\dfrac{2}{x}\right)\,dx \\&=\left(-5x^{-3}+2\ln(x)\right)\Bigg|^{3}_{{1}}\end{aligned}$ Second, plug in the limits of integration: $(2\ln({3})-5\cdot{3}^{-3})-(2\ln({1})-5\cdot{1}^{-3}) = 2\ln\left(3\right)+\dfrac{130}{27}$. The answer: $\int^{3}_{1}\left(\dfrac{15+2x^3}{x^4}\right)\,dx ~=~2\ln\left(3\right)+\dfrac{130}{27}$